Riddle: A storm rains an average of 0.8 inches over an area of 135 square miles. How much latent heat of condensation was produced by the storm in order to produce this mass of rainfall? Final answer will be in Joules. Water density = 1 g/cm^3.






Answer to Riddle: In order to know how much Energy is produced we need to determine the mass of the rain that fell. Density = mass / volume. We know the density of water and we can find the volume of the water. Once we find the volume of the water we can solve the density formula for mass.

Volume of rain will be the area the rain fell on times the depth of the rain. The depth of the rain is 0.8 inches. In order to get the final answer in Joules we need to convert to the metric system for the area the rain fell on. Square miles will need to be converted the square centimeters and the rain depth will need to be in centimeters.

The following conversion factors are need to convert square miles to square centimeters:

1 mile = 5,280 feet
12 inches = 1 foot
1 inch = 2.54 centimeters

The conversion may end up the most challenging aspect of finding the answer. Units must be cancelled properly and if units are squared then the number must be squared also.

First convert 135 square miles to square feet

135 mi^2 * (5,280^2 ft^2 / 1 mi^2) = 3,763,584,000 ft^2

Next convert ft^2 to in^2

3,763,584,000 ft^2 * (12^2 in^2 / 1 ft^2) = 5.41956096 * 10^11 in^2

Now convert in^2 to cm^2 to get it into metric units

5.41956096 * 10^11 in^2 * (2.54^2 cm^2 / 1 in^2) = 3.496483949 * 10^12 cm^2

Since we have the area the rain fell over in square centimeters we can multiply by the rain depth to get the volume of rain in cm^3.

0.8 in (2.54 cm / 1 in) = 2.032 cm

Volume of rain = 3.496483949 * 10^12 cm^2 * 2.032 cm = 7.104855384 * 10^12 cm^3

The density formula can now be used to determine the mass of rain

Density = mass / volume

1 g/cm^3 = mass / 7.104855384 * 10^12 cm^3

mass = 7.104855384 * 10^12 cm^3 * 1 g/cm^3 = 7.104855384 * 10^12 grams

The latent heat of condensation has a known value of 2.5 * 10^6 Joules/kg

convert mass to kg since latent heat has kg in units

(7.104855384 * 10^12 grams) * (1 kg / 1,000 g) = 7.104855384 * 10^9 kg

Total energy is equal to mass of rain times the latent heat of condensation

2.5 * 10^6 Joules/kg * 7.104855384 * 10^9 kg = 1.776213846 * 10^16 Joules

written out and rounded this is 17,700,000,000,000,000 Joules

This is 17.7 quadrillion Joules which is an amazing amount of energy just from the condensation produced from a rain complex.

謎語(yǔ)：暴風(fēng)雨在 0.8 平方英里的區(qū)域內(nèi)平均降雨量為 135 英寸。如何 風(fēng)暴產(chǎn)生了大量的冷凝潛熱，以產(chǎn)生這種潛熱 降雨量？最終答案將以焦耳為單位。水密度 = 1 g/cm^3。






謎語(yǔ)的答案：為了知道產(chǎn)生了多少能量，我們需要確定 落下的雨水。密度=質(zhì)量/體積。我們知道水的密度和 我們可以找到水的體積。一旦我們找到了水的體積，我們就可以解決 質(zhì)量的密度公式。

降雨量將是降雨面積乘以降雨深度。深度 降雨量為 0.8 英寸。為了得到焦耳的最終答案，我們需要轉(zhuǎn)換為 降雨區(qū)域的公制。平方英里將需要轉(zhuǎn)換 平方厘米和雨深需要以厘米為單位。

將平方英里轉(zhuǎn)換為平方厘米需要以下轉(zhuǎn)換系數(shù)：

1 英里 = 5,280 英尺 12 英寸 = 1 英尺

1 英寸 = 2.54 厘米

轉(zhuǎn)換可能最終成為尋找答案最具挑戰(zhàn)性的方面。單位必須 正確取消，如果單位是平方，則數(shù)字必須 平方也。

首先將 135 平方英里轉(zhuǎn)換為平方英尺

135 mi^2 * （5,280^2 ft^2 / 1 mi^2） = 3,763,584,000 ft^2
下一頁(yè) 將 ft^2 轉(zhuǎn)換為 in^2
3,763,584,000 ft^2 * （12^2 in^2 / 1 ft^2） = 5.41956096 * 10^11 in^2
現(xiàn)在將 in^2 轉(zhuǎn)換為 cm^2 以將其轉(zhuǎn)換為公制單位

5.41956096 * 10^11 in^2 * （2.54^2 cm^2 / 1 in^2） = 3.496483949 * 10^12 cm^2




由于我們有降雨量以平方厘米為單位的面積，我們可以乘以 通過(guò)降雨深度得到以 cm^3 為單位的雨量。

0.8 英寸（2.54 厘米/1 英寸）= 2.032 厘米

雨量 = 3.496483949 * 10^12 cm^2 * 2.032 cm = 7.104855384 * 10^12 cm^3 密度公式現(xiàn)在可用于確定雨

的質(zhì)量 密度 = 質(zhì)量 /
體積 1 g/cm^3 = 質(zhì)量 / 7.104855384 * 10^12 cm^3
質(zhì)量 = 7.104855384 * 10^12 cm^3 * 1 g/
cm^3


= 7.104855384 * 10^12 克

冷凝潛熱的已知值為 2.5 * 10^6 焦耳/千克，將質(zhì)量轉(zhuǎn)換為千克，因?yàn)闈摕岬膯挝?br _istranslated="1"/>
為 kg （7.104855384 * 10^12 克） * （1 千克/1,000 克） = 7.104855384 * 10^9 千克 總能量等于雨的質(zhì)量乘以冷凝
潛熱
2.5 * 10^6 焦耳/千克 * 7.104855384 * 10^9 千克



= 1.776213846 * 10^16 焦耳

寫出來(lái)并四舍五入，這是 17,700,000,000,000,000 焦耳
這是 17.7 萬(wàn)億焦耳
，這是一個(gè)驚人的能量量 由雨水復(fù)合物產(chǎn)生的冷凝水。